2025 AMC 10A 第 7 题

Question

设

a

和

b

是实数。多项式

x^3 + x^2 + ax + b

除以

x - 1

时余数为

4

，除以

x - 2

时余数为

6

。

b - a

等于多少？

Suppose

a

and

b

are real numbers. When the polynomial

x^3 + x^2 + ax + b

is divided by

x - 1,

the remainder is

4.

When the polynomial is divided by

x - 2,

the remainder is

6.

What is

b - a?

14

15

16

17

18

答案：E

知识点：多项式方程组

难度评级：1250

小提示：

根据余数定理，分别在

x = 1

和

x = 2

处代入多项式。

By the Remainder Theorem, evaluate the polynomial at

x = 1

and

x = 2

大提示：

这给出

a + b + 2 = 4

和

2a + b + 12 = 6

。

This gives

a + b + 2 = 4

and

2a + b + 12 = 6

解答：

由余数定理，直接代入即可。我们有

p(1) = 1 + 1 + a + b = 4

，所以

a + b = 2

。又有

p(2) = 8 + 4 + 2a + b = 6

，所以

2a + b = -6

。第二式减去第一式，得

a = -8

，从而

b = 10

。因此

b - a = 10 - (-8) = 18

，所以正确答案是 E。

By the Remainder Theorem, just plug in. We get

p(1) = 1 + 1 + a + b = 4,

so

a + b = 2.

And

p(2) = 8 + 4 + 2a + b = 6,

so

2a + b = -6.

Subtract the first from the second:

a = -8,

hence

b = 10.

Then

b - a = 10 - (-8) = 18.

Thus, E is the correct answer.

2025 AMC 10A 第 7 题

答案：E

难度评级：1250

小提示：

大提示：

解答：

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